Assume that $k$ is a field and that $k(X)$ is the field of rational functions in $X$ with coefficients in $k$. If $f(Y)$ and $g(Y)$ are coprime polynomials in $k[Y]$, then $f(Y)-g(Y)X$ must be irreducible in $k(X)[Y]$.
Not sure how to approach this. Would proving that $f(Y)-g(Y)X$ is irreducible in $k[X][Y]$ help? I was hoping to use some form of Gauss's Lemma and prove that irreducibility in $k[X][Y]$ implies irreduciblity in $k(X)[Y].$ If $f(Y)-g(Y)X$ factors in $k[X][Y]$ it seems to me (considering the degree of $X$) that one factor must be a polynomial in $Y$ alone. But not sure how to use these facts! Not sure about this argument but here goes. Assume that $f(Y)-g(Y)X = h(Y)(r(Y)+s(Y)X)=h(Y)r(Y)+h(Y)s(Y)X$, then comparing like powers of $X$ in these two factorizations we can conclude that $f(Y) = h(Y)r(Y)$ and $g(Y)=h(Y)s(Y)$ but this forces us to conclude $h(Y)$ must be a unit since by hypothesis $f(Y)$ and $g(Y)$ are coprime. Hence $f(Y)-g(Y)X$ must be irreducible in $k[X][Y]$. Now I wonder of I can use Gauss's Lemma to prove that this polynomial remains irreducible in $k(X)[Y]$.
The discussion above proves that $f(Y)-g(Y)X$ is irreducible in $k[X][Y]$. Assume that we have a factorization $$(f(Y)-g(Y)X)={h_1}(X,Y){h_2}(X,Y)$$ where ${h_1}(X,Y)$, and ${h_2}(X,Y)$ are polynomials in $k(X)[Y]$. Multiplying both sides of the above displayed identity by the least common multiple of the denominators in the coefficients of the powers of $Y$ in ${h_1}(X,Y)$, and ${h_2}(X,Y)$, we have a factorization $${\lambda}(X)(f(Y)-g(Y)X)={h'_1}(X,Y){h'_2}(X,Y)$$ where ${\lambda}(X)\in{k[X]}$, and ${h'_1}(X,Y)$, and ${h'_1}(X,Y)$ are both polynomials in $k[X][Y].$ We can now prove by contradiction (exactly the same argument in the classical Gauss Lemma involving $\bf{Z}$ and $\bf{Q}$) that every irreducible (i.e. prime) factor of ${\lambda}(X)$ either divides $\it{every}$ coefficient of ${h'_1}(X,Y)$, or divides $\it{every}$ coefficient of ${h'_2}(X,Y).$ Thus every irreducible factor of ${\lambda}(X)$ can be divided out (and hence ${\lambda}(X)$ itself may be divided out yielding a factorization $$(f(Y)-g(Y)X)={h'}(X,Y){h''}(X,Y)$$ in $k[X][Y]$. This contradicts our observation that $(f(Y)-g(Y)X)$ is irreducible in $k[X][Y]$.